Question 532499
I'm looking at this from my way, which may not be the way you were taught, so I hope you can follow.

I'm going to be using the formula {{{Distance = Rate*Time}}} for this, and modifying it as needed.

First, their rates {{{R=D/T}}} 

Roger's Rate: {{{1 mile/8 min}}} and Jeff's Rate: {{{1 mile/6 min}}}

Now, their times for this race {{{T=D/R}}} We don't know the distance, so for Roger: {{{T=D/(1mile/8min)}}}. For Jeff, it's a little more complicated, because we have to add on that extra 1 min: {{{T=1 min + D/(1mile/6min)}}}

Both of these formulas can be simplified because we have fractions in fractions to get:

Roger: {{{T=8D}}} and Jeff: {{{T=1+6D}}}

The same amount of time will have elapsed when Jeff catches Roger, so set the 2 equations equal to each other, and solve for the distance.

Another way to do this is to graph both of their movements, having time be one axis and distance be the other. You will be able to see where the two intersect.